The adjoint problem

The the study of bifurcation below, it will be of interest to consider the inhomogeneous problem  

$$(A-\sigma I)X=F,$$ (10)

where $\sigma$ is an eigenvalue of A. If $\sigma$ is an eigenvalue of geometric multiplicity k, then F must satisfy k solvability conditions for a solution of (1.11) to exist. These solvability conditions can be phrased in terms of the adjoint problem. Let $(\cdot,\cdot)$ denote the usual inner product:  

$$(Y,X)=\sum_{i=1}^n \bar Y_iX_i.$$ (11)

Suppose now that b is orthogonal to every F for which (1.11) has a solution. Then  

$$(b,(A-\sigma I)X)=((A^*-\bar\sigma I)b,X)=0$$ (12)

for every X. Here A^* denotes the adjoint matrix: $A^*_{ij}=\bar A_{ji}$. We conclude from (1.13) that $(A^*-\bar\sigma I)b=0$, i.e. b is an eigenvector of A^* with eigenvalue $\bar\sigma$. The equation (1.11) is solvable if and only if F is orthogonal to every such eigenvector b.

Example 4

 The matrix

$$A=\pmatrix{0&0\cr 1&1}$$ (13)

has a simple eigenvalue 0 with eigenvector (1,-1)^T. The adjoint matrix is

$$A^*=\pmatrix{0&1\cr 0&1},$$ (14)

with eigenvector b=(1,0)^T. The system

   $$\begin{eqnarray} 0x_1+0x_2&=&f_1,\nonumber\ x_1+x_2&=&f_2.\end{eqnarray}$$

is solvable if an only if (b,F)=f₁=0.

By a general theorem of linear algebra, the codimension of the range of $A-\sigma I$ must equal the dimension of its nullspace. Hence the geometric multiplicity of the adjoint eigenvalue problem is equal to that of the original eigenvalue problem. Let $\sigma$ be a specific eigenvalue A with geometric multiplicity k and let $\xi_1,\xi_2,...,\xi_k$ be k linearly independent eigenvectors. Then there exist k linearly independent eigenvectors b₁,b₂,...,b_k of the adjoint matrix A^* with eigenvalue $\bar\sigma$. It can be shown that if the algebraic multiplicity of the eigenvalue is also equal to k, then the $k\times k$ matrix $M_{ij}=(b_i,\xi_j)$ is nonsingular. We can then replace the b_i by their linear combinations

$$\tilde b_i=\sum_{m=1}^k \bar M^{-1}_{im}b_m.$$ (15)

We check that  

$$(\tilde b_i,\xi_j)=\sum_{m=1}^k M^{-1}_{im}(b_m,\xi_j) =\sum_{m=1}^k M^{-1}_{im}M_{mj}=\delta_{ij}.$$ (16)

We shall now assume that the b_i have in the first place been chosen such that the normalization (1.18) holds and drop the tilde.

We can now define a projection operator  

$$PF=F-\sum_{i=1}^k (b_i,F)\xi_i.$$ (17)

It is easily checked that P(PF)=PF and that (b_i,PF)=0 for every i. That is, PF lies in the range of $A-\sigma I$. The equation (1.19) thus provides a linear decomposition of F into PF and a linear combination of the eigenvectors $\xi_i$. We refer to P as a projection operator; for any F, PF is the part of F in the decomposition which lies in the range of $A-\sigma I$. The equation $(A-\sigma I)X=F$ can be decomposed into two parts using the projection P. Let Y=PX, Z=(I-P)X, G=PF, H=(I-P)F. Then we find

$$(A-\sigma I)Y=G,\quad (A-\sigma I)Z=H.$$ (18)

The first of these equations can be solved uniquely for Y as a function of G. The second equation simplifies to H=0, since $(A-\sigma I)Z=0$. Hence we have the solvability condition H=0, and Z is arbitrary.